Ngô Quốc Anh

April 7, 2008

Integral with fraction of Min and Max of a function

Filed under: Giải Tích 2 — Ngô Quốc Anh @ 2:37

Let be a positive non-zero integer. Prove that:

\int_0^1 \frac {\min \left(x^n,(1 - x)^n\right)}{\max \left(x^n,(1 - x)^n\right)} \, dx = 2 n ( - 1)^{n - 1} \left(\log (2) + \sum _{i = 1}^{n - 1} \frac {( - 1)^i}{i}\right) - 1.

Hint. Simplify the integral to:

\int_0^1 \frac {\min \left(x^n,(1 - x)^n\right)}{\max \left(x^n,(1 - x)^n\right)} \, dx = \int_0^{\frac {1}{2}} \left(\frac {1}{x} - 1\right)^{ - n} \, dx + \int_{\frac {1}{2}}^1 \left(\frac {1}{x} - 1\right)^n \, dx.

Use the Symmetry

\int_0^{\frac {1}{2}} \left(\frac {1}{x} - 1\right)^{ - n} \, dx = \int_{\frac {1}{2}}^1 \left(\frac {1}{x} - 1\right)^n \, dx.

Use newton

\left(\frac {1}{x} - 1\right)^n = ( - 1)^n + \frac {( - 1)^{n - 1} n}{x} + \sum _{i = 2}^n x^{ - i} ( - 1)^{n - i} \left( \begin{array}{c} n \\ i \end{array} \right).

Integrate the last sum and prove that:

\sum _{i = 2}^n \frac {( - 1)^{n - i} \left(2^i - 2\right) }{2 (i - 1)}\left( \begin{array}{c} n \\ i \end{array} \right) = n ( - 1)^{n - 1} \sum _{i = 1}^{n - 1} \frac {( - 1)^i}{i} - \frac{1}{2} \left((-1)^n+1\right).

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